Proof of Implicit Differentiation (showing a statement is true) 
Prove that if $x^2+y^2-2y\sqrt{1+x^2} = 0$, then $dy/dx = x/\sqrt{1+x^2}$.

Whilst I have implicitly differentiated in terms of x in order to derive that
$$dy/dx = (-x+2xy/\sqrt{1+x^2})/(y\sqrt{1+x^2}-1-x^2)$$
however I am unsure as to what my next steps are. I believe that I will need to rearrange the original supposition in order to achieve the proof however I do require some help as to how can I do this as I cannot see what would possibly cancel out.
Additionally, does anyone have any particularly handy methods/techniques to be able to complete proofs of this format of question more easily? I do understand that there  is not one singular technique that can be used when completing this proofs but is there any strategy that minimizes the number of dead-ends that I come across in my working?
Thanks
 A: A simpler way: the identity
$$x^2+y^2-2y\sqrt{1+x^2} = 0$$
is equivalent to
$$y^2+(1+x^2)-2y\sqrt{1+x^2} = 1$$
that is
$$(y-\sqrt{1+x^2})^2=1$$
and differentiating  in terms of $x$ both sides, we obtain
$$2(y-\sqrt{1+x^2})\left(y'-\frac{x}{\sqrt{1+x^2}}\right)=0.$$
The first factor is always different from zero: in fact letting $y=\sqrt{1+x^2}$ into the given equation we get
$$x^2+(1+x^2)-2(1+x^2) = 0$$
that is $-1=0$ which is impossible. Therefore the second factor has to be zero, and we find
$$y'=\frac{x}{\sqrt{1+x^2}}.$$
Preliminary step. Notice that by the two dimensional implicit function theorem, $y'(x)$ exists (and therefore we are allowed to take the derivative in terms of $x$) if
$$\frac{\partial F}{\partial y}=2(y-\sqrt{1+x^2})\not=0$$
where $F(x,y)=x^2+y^2-2y\sqrt{1+x^2}$, which holds as shown above.
A: Solve the quadratic of $y$ to get $$u=\sqrt{1+x^2}\pm 1 \implies \frac{dy}{dx}=
\frac{x}{\sqrt{1-x^2}}.$$
A: I've a short trick for this.
For an implicit equation of $x$ and $y$ such that $f(x, y) = 0$, following result holds true:
$$ \dfrac{dy}{dx} = \dfrac{-\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$$
For the given equation $x^2 + y^2 - 2y\sqrt{1+x^2} = 0$, derivative of $y$ with respect to $x$ is given by,
$$\begin{aligned}\dfrac{dy}{dx} &= -\left[\dfrac{\frac{\partial }{\partial x}(x^2 + y^2 - 2y\sqrt{1+x^2} )}{\frac{\partial }{\partial y}(x^2 + y^2 - 2y\sqrt{1+x^2} )}\right]\\& = -\left[\dfrac{2x - 2y\cdot \frac{x}{\sqrt{1+x^2}}}{2y - 2\sqrt{1+x^2} }\right]\\& =-\left[\dfrac{x(\sqrt{1+x^2} - y)\cdot\frac1{\sqrt{1+x^2}}}{(y - \sqrt{1+x^2} )}\right] \\& =\frac{x}{\sqrt{1+x^2}}\end{aligned}$$
A: We find from the given relation $ \ x^2 + y^2 \ = \  2y\sqrt{1+x^2}  \ $  that
$$ 2x \ + \ 2y·\frac{dy}{dx} \ \ =  \ \ 2·\frac{dy}{dx}·\sqrt{1+x^2} \ + \ 2y·\frac12·\frac{1}{\sqrt{1+x^2}}·2x \ $$
$$ \Rightarrow \ \   ( \ y \ - \ \sqrt{1+x^2} \ ) ·\frac{dy}{dx} \ \ =  \ \     y· \frac{1}{\sqrt{1+x^2}}· x \ - \ x \ \ \Rightarrow \ \  \frac{dy}{dx} \ \ = \ \   \frac{-x \ + \  \frac{xy}{\sqrt{1+x^2}}}{y \ - \ \sqrt{1+x^2} } \ \ . $$
Replacing (at least for the present) the radical using the original equation $ ( \ \sqrt{1+x^2} \ = \ \frac{x^2 + y^2}{2y} \ ) \ $ , we obtain
$$  \frac{dy}{dx} \ \ = \ \   \frac{  xy·\frac{2y}{x^2 + y^2} \ - \ x}{y \ - \ \frac{x^2 + y^2}{2y} } \ \ = \ \  \frac{  x·\left( \  2y^2  \ - \ [x^2 + y^2] \ \right)}{y·(x^2 + y^2) \ - \ \frac{(x^2 + y^2)^2}{2y} } \ \ = \ \ \frac{  x· (    y^2  \ - \  x^2      )·2y}{2y^2·(x^2 + y^2) \ - \  (x^2 + y^2)^2  }  $$
$$ = \ \ \frac{  x· (    y^2  \ - \  x^2      )·2y}{(2y^2 \ - \ x^2 \ - \ y^2)·(x^2 + y^2)   } \ \ = \ \ \frac{  x· (    y^2  \ - \  x^2      )·2y}{( y^2 \ - \ x^2 )·(x^2 + y^2)   } \ \ = \ \ x· \frac{  2y}{  x^2 + y^2    } 
$$
$$= \ \ x· \frac{1}{ \sqrt{1+x^2}   } \ \ ,  $$
again using the original relation.  (So it appears there may be an error in your expression for the implicit derivative.)
